Exam #1 Answer Key
Economics 435: Quantitative Methods
Spring 2005
1
Regression with standardized variables
a)
β̂0 = 0
b)
R2 = β̂12
c)
corr(ỹ,
ˆ
x̃) = β̂1
d) Well, since we have shown that βˆ1 = corr(x̃,
ˆ
ỹ), and all correlations are between −1 and +1, the answer
to the question is “no.”
2
Heteroskedasticity: A simple example
a) There are many ways to show this, but here’s one:
β̂1
=
=
=
=
=
=
=
cov(y,
ˆ
x)
var(x)
ˆ
Pn
1/n i=1 (yi − ȳ)(xi − x̄)
Pn
1/n i=1 (xi − x̄)2
Pn
Pn
i=1 yi (xi − x̄) − ȳ
i=1 (xi − x̄)
nx̄(1 − x̄)
Pn
x
y
(1
− x̄) + (1 − xi )yi (0 − x̄)
i=1 i i
nx̄(1 − x̄)
Pn
Pn
(1 − x̄) i=1 xi yi − x̄ i=1 (1 − xi )yi
nx̄(1 − x̄)
Pn
Pn
x
y
(1 − xi )yi
i=1 i i
− i=1
nx̄
n(1 − x̄)
ȳ1 − ȳ0
For β̂0 , we simply note that ȳ = (1 − x̄)ȳ0 + x̄ȳ1 , and substitute.
b) Assumptions SLR1-SLR4 are met, so the OLS estimators are unbiased and consistent.
1
ECON 435, Spring 2005
2
c)
var(β̂0 )
= var(ȳ0 )
σ02
=
n(1 − x̄)
var(β̂1 )
= var(ȳ1 ) + var(ȳ0 ) − 2cov(ȳ1 , ȳ0 )
σ12
σ02
=
+
nx̄ n(1 − x̄)
cov(β̂0 , β̂1 )
3
= cov(ȳ0 , ȳ1 − ȳ0 )
= cov(ȳ0 , ȳ1 ) − var(ȳ0 )
σ02
=
n(1 − x̄)
Hypothesis testing
a) Our best chance for rejecting the null is if all of our observations are
√ greater than zero (ȳn = 1) of if all
observations are less than zero√(ȳn = 0). If this is the case then |t| = n. With a sample size of two, the
largest
possible value for |t| is 2 = 1.414 < 1.96. In general, we will have no chance of rejecting the null if
√
n < 1.96 or if n < 1.962 = 3.84. In other words we need a sample of size four or greater to have a chance
at rejecting the null.
b) I would define:
t=
n
X
I(xi > 10)
i=1
Under the null hypothesis:
Pr(t = 0) = 1
So we can reject the null at the 5% significance level (actually, at any significance level) if t > 0.
In other words, we can reject the null if we ever see a value of x bigger than 10, and we cannot reject the
null otherwise. This makes sense, hopefully.
4
Craps
a)
Pr(x = 2)
Pr(x = 3)
Pr(x = 4)
Pr(x = 5)
Pr(x = 6)
Pr(x = 7)
Pr(x = 8)
Pr(x = 9)
Pr(x = 10)
Pr(x = 11)
Pr(x = 12)
=
=
=
=
=
=
=
=
=
=
=
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
ECON 435, Spring 2005
3
b) The probabilities are:
Pr(x ∈ {7, 11})
Pr(x ∈ {2, 3, 12})
Pr(x = 4)
Pr(x = 5)
Pr(x = 6)
Pr(x = 8)
Pr(x = 9)
Pr(x = 10)
=
=
=
=
=
=
=
=
8/36
4/36
3/36
4/36
5/36
5/36
4/36
3/36
c) The probabilities are:
Pr(x = 4|x ∈ {4, 7})
=
Pr(x = 5|x ∈ {5, 7})
=
Pr(x = 6|x ∈ {6, 7})
=
Pr(x = 8|x ∈ {8, 7})
=
Pr(x = 9|x ∈ {9, 7})
=
Pr(x = 10|x ∈ {10, 7})
=
3/36
3/36 + 6/36
4/36
4/36 + 6/36
5/36
5/36 + 6/36
5/36
5/36 + 6/36
4/36
4/36 + 6/36
3/36
3/36 + 6/36
= 1/3
= 2/5
= 5/11
= 5/11
= 2/5
= 1/3
d) The probability is:
8/36 + 3/36 ∗ 1/3 + 4/36 ∗ 2/5 + 5/36 ∗ 5/11 + 5/36 ∗ 5/11 + 4/36 ∗ 2/5 + 3/36 ∗ 1/3
244
=
495
≈ 49.3%
Pr(win) =
e) For every $1 bet, the casino’s profit is $1 with 50.7% probability and -$1 with 49.3% probability. The
expected profit per $1 bet is thus approximately 1.4 cents.